Question: What is the extraneous solution to these equations? $\dfrac{x^2 + 6}{x - 8} = \dfrac{9x - 2}{x - 8}$
Explanation: Multiply both sides by $x - 8$ $ \dfrac{x^2 + 6}{x - 8} (x - 8) = \dfrac{9x - 2}{x - 8} (x - 8)$ $ x^2 + 6 = 9x - 2$ Subtract $9x - 2$ from both sides: $ x^2 + 6 - (9x - 2) = 9x - 2 - (9x - 2)$ $ x^2 + 6 - 9x + 2 = 0$ $ x^2 + 8 - 9x = 0$ Factor the expression: $ (x - 8)(x - 1) = 0$ Therefore $x = 8$ or $x = 1$ At $x = 8$ , the denominator of the original expression is 0. Since the expression is undefined at $x = 8$, it is an extraneous solution.